Andy has a certain number of $1 bills, $2 bills and $10 bills. If the

Question

Andy has a certain number of $1 bills, $2 bills and $10 bills. If the number of $1 bills he has is six times the number of $2 bills he has, and the total amount he has is $160, what is the maximum number of $10 bills, Andy could have?

A. 12

B. 10

C. 8

D. 6

E. 5

A. 12

B. 10

C. 8

D. 6

E. 5

randomavoidplease · Answer

I took the basic approach.

1x + 2y + 10k = 160$ 
X, Y & K being the number of bills of the respective denomination.

Also  x = 6y

Therefore  8y + 10k = 160$ 
Factoring out 2, gives us  4y + 5k = 80$

Now checking from the top if K = 12 then he has 60$ and  4y = 20$ , a multiple of 4. So 12 bills of $10 is the maximum possibility.

Mona2019 · Answer

Let the number of 2$ bill be X,
Number of 1$ bill is 6X
Total value of these two becomes 8X
As the total amount is 160, and the denomination left is 10$ bill, so 8x should be a multiple of 10,
So that we can have an integer value for number of 10$ bills, 
Minimum Value X can take is 5 , so that 8x = 40
So 10y= 120
Y= 12
So A , IMO is the answer

Posted from my mobile device

ScottTargetTestPrep · Answer

Solution:
 
If we let x = the number of $2 bills Andy has, then 6x = the number of $1 bills he has. Furthermore, since we are looking for the maximum number of $10 bills Andy could have, let’s check the answer choices starting with the largest number given, which is 12. If he has 12 $10 bills, then he has $40 left for the $1 bills and $2 bills. We can create the equation:

1(6x) + 2(x) = 40

8x = 40

x = 5

Since x is an integer, we see that it is possible for Andy to have 12 $10 bills (note: the only way we can reject 12 as the correct answer is if we can show x, the number of $2 bills, is not a positive integer).

Alternate Solution:

If we let x = the number of $2 bills Andy has, then 6x = the number of $1 bills he has. The dollar value of the $1 bills is 6x dollars and the dollar value of the $2 bills is 2x dollars. Together, the $1 and $2 bills make up 6x + 2x = 8x dollars. Thus, the $10 bills are worth 160 - 8x dollars and it follows that the number of $10 bills is (160 - 8x)/10. Since the number of $10 bills must be an integer, it must be true that (160 - 8x)/10 is an integer. To maximize the expression (160 - 8x)/10, we must minimize the negative term, 8x. The smallest non-zero value of x for which 160 - 8x is divisible by 10 is x = 5. Thus, the greatest value of $10 bills is (160 - 8x)/10 = (160 - 8(5))/10 = (160 - 40)/10 = 120/10 = 12.

Answer: A

MathRevolution · Answer

If there is 1 bill of $2 then there will be 6 bills for $1.

Total: $1(6) + $2(1) = 8.

This pair can give us a $40 amount as a minimum so that the remaining amount $160- $40 = $120 is divisible by $10.

So, 12 * 10 = 120

Answer A

DhanyaAbhirami · Answer

The question doesn't mention non-zero number of dollar bills of any denomination. So it can be 16 bills of 10 dollar denomination also. 
Am I missing something?

Bunuel · Answer

If the number of $1 bills and $2 bills were both 0, while it would technically be correct to say that the number of $1 bills is six times the number of $2 bills, it wouldn’t make much practical sense in the context of the problem.

bumpbot · Answer

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